The sum of the series `1/(sqrt2 + sqrt1)` + `1/(sqrt2 + sqrt3)` + ……+ `1/ (sqrt120 + sqrt121)` is .

Options

A. 10
B. 11
C. 12
D. none of these

Answer

To solve the series \( S = \frac{1}{\sqrt{2} + \sqrt{1}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \ldots + \frac{1}{\sqrt{120} + \sqrt{121}} \), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term of the series can be expressed as:
\[ T_n = \frac{1}{\sqrt{n} + \sqrt{n-1}} \]
where \( n \) ranges from 2 to 121.

2. **Rationalize the Denominator**:
To simplify \( T_n \), we will rationalize the denominator:
\[ T_n = \frac{1}{\sqrt{n} + \sqrt{n-1}} \cdot \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n} - \sqrt{n-1}} = \frac{\sqrt{n} - \sqrt{n-1}}{(\sqrt{n})^2 - (\sqrt{n-1})^2} \]
The denominator simplifies to:
\[ n - (n-1) = 1 \]
Thus, we have:
\[ T_n = \sqrt{n} - \sqrt{n-1} \]

3. **Write the Series in Summation Form**:
The series can now be rewritten as:
\[ S = \sum_{n=2}^{121} (\sqrt{n} - \sqrt{n-1}) \]

4. **Apply the Telescoping Property**:
This series is telescoping, meaning most terms will cancel out:
\[ S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \ldots + (\sqrt{121} - \sqrt{120}) \]
When we expand this, we see that all intermediate terms cancel:
\[ S = \sqrt{121} - \sqrt{1} \]

5. **Calculate the Final Result**:
Now we can compute the remaining terms:
\[ S = 11 - 1 = 10 \]

### Final Answer:
The sum of the series is \( \boxed{10} \).